An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Figures 1-4 illustrate the rotation rules.

Now given a sequence of insertions, you are supposed to tell the root of the resulting AVL tree.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=20) which is the total number of keys to be inserted. Then N distinct integer keys are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print ythe root of the resulting AVL tree in one line.

Sample Input 1:

5
88 70 61 96 120

Sample Output 1:

70

Sample Input 2:

7
88 70 61 96 120 90 65

Sample Output 2:

88
 #include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std; struct node
{
node(int v):left(NULL),right(NULL),high(),val(v){}
node* left,* right;
int val,high;
}; int gethigh(node* root)
{
int a = , b = ;
if(root->left!= NULL)
a = root->left->high;
if(root->right!= NULL)
b = root->right->high;
return a > b ? a+:b+;
} void R(node* & root)
{
node* tem = root->left;
root->left = tem->right;
tem->right = root;
root->high = gethigh(root);
tem->high = gethigh(tem);
root = tem;
} void L(node* & root)
{
node* tem = root->right;
root->right = tem->left;
tem->left = root;
root->high = gethigh(root);
tem->high = gethigh(tem);
root = tem;
} void insert(node*& root,int val)
{
if(root == NULL)
{
root = new node(val);
return;
} if(val < root->val)
{
insert(root->left,val);
root->high = gethigh(root);
int a = root->left == NULL ? : root->left->high;
int b = root->right == NULL ? : root->right->high;
if(a - b == )
{
int c = root->left->left == NULL ? :root->left->left->high;
int d = root->left->right == NULL ? :root->left->right->high;
if(c - d == )
{
R(root);
}
else if(c - d == -)
{
L(root->left);
R(root);
}
}
}
else
{
insert(root->right,val);
root->high = gethigh(root);
int a = root->left == NULL ? : root->left->high;
int b = root->right == NULL ? : root->right->high;
if(a - b == -)
{
int c = root->right->right == NULL ? :root->right->right->high;
int d = root->right->left == NULL ? :root->right->left->high;
if(c - d == )
{
L(root);
}
else if(c - d == -)
{
R(root->right);
L(root);
}
}
}
} int main()
{
int n,tem;
scanf("%d",&n);
node* Tree = NULL;
for(int i = ;i < n;++i)
{
scanf("%d",&tem);
insert(Tree,tem);
}
printf("%d\n",Tree->val);
return ;
}

1066. Root of AVL Tree (25)的更多相关文章

  1. pat 甲级 1066. Root of AVL Tree (25)

    1066. Root of AVL Tree (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue An A ...

  2. PAT 甲级 1066 Root of AVL Tree (25 分)(快速掌握平衡二叉树的旋转,内含代码和注解)***

    1066 Root of AVL Tree (25 分)   An AVL tree is a self-balancing binary search tree. In an AVL tree, t ...

  3. PAT甲级:1066 Root of AVL Tree (25分)

    PAT甲级:1066 Root of AVL Tree (25分) 题干 An AVL tree is a self-balancing binary search tree. In an AVL t ...

  4. PAT 1066. Root of AVL Tree (25)

    An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child sub ...

  5. 1066 Root of AVL Tree (25)

    An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child sub ...

  6. PAT Advanced 1066 Root of AVL Tree (25) [平衡⼆叉树(AVL树)]

    题目 An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child ...

  7. 1066 Root of AVL Tree (25分)(AVL树的实现)

    An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child sub ...

  8. PAT (Advanced Level) 1066. Root of AVL Tree (25)

    AVL树的旋转.居然1A了.... 了解旋转方式之后,数据较小可以当做模拟写. #include<cstdio> #include<cstring> #include<c ...

  9. PAT甲级题解-1066. Root of AVL Tree (25)-AVL树模板题

    博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6803291.html特别不喜欢那些随便转载别人的原创文章又不给 ...

随机推荐

  1. Android小项目练习之一 项目简介

    ------- 源自梦想.永远是你IT事业的好友.只是勇敢地说出我学到! ---------- 按惯例,写在前面的:可能在学习Android的过程中,大家会和我一样,学习过大量的基础知识,很多的知识点 ...

  2. HDU 1257 最少拦截系统 (DP || 贪心)

    最少拦截系统 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Statu ...

  3. 配置多个Log4net实例

    应用场景: 控制软件工程拆分为多个子项目,在开发阶段希望每个子项目的日志是单独的.同时又希望日志统一集中定义在Tools(Tools为工具类项目,生成Tools.dll)下,而不是分散在各个子项目中. ...

  4. 剑指Offer35 两个链表第一个公共结点

    /************************************************************************* > File Name: 35_FirstC ...

  5. jQuery选择器之层次选择器Demo

    测试代码: 02-层次选择器.html <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" &quo ...

  6. 自定义的UITabbar上面的按钮的x坐标的计算方法

    ; i < 4; i++) {//4是按钮的个数 NSString *backImage = backgroud[i]; NSString *heightImage = heightBackgr ...

  7. css3 text-transform变形动画

    详细内容请点击 版本:CSS1 兼容性:IE4+ NS4+ 继承性:有 语法: text-transform : none | capitalize| uppercase| lowercase 参数: ...

  8. sql语句使用游标修改表中数据

    declare @a varchar(),@b varchar() declare user_cursor cursor for select a,b from tableA tab open use ...

  9. Android文字转语音

    虽然视觉上的反馈通常是给用户提供信息最快的方式,但这要求用户把注意力设备上.当用户不能查看设备时,则需要一些其他通信的方法.Android提供了强大的文字转语音Text-to-Speech,TTS A ...

  10. Js中把JSON字符串转换为JSON对象(eval()、new Function())

    在JS中将JSON的字符串解析成JSON数据格式,一般有两种方式: 1.一种为使用eval()函数. 2. 使用Function对象来进行返回解析. 第一种解析方式:使用eval函数来解析,并且使用j ...